# Solve for junction capacitance from diode's SPICE parameters

I am trying to calculate maximum junction capacitance (or depletion capacitance) $$\C_j\$$ for the BAT62-02L diode from this datasheet based on its SPICE parameters:

• $$\I_s=\$$ IS $$\=\$$ Saturation current $$\=250~\text{nA}\$$.
• $$\\eta=\$$ XN $$\=\$$ N $$\=\$$ Emission coefficient $$\=1.04\$$.
• $$\r_s=\$$ RS $$\=\$$ Series resistance $$\=190~\Omega\$$.
• $$\C_{j0}=\$$ CJ0 $$\=\$$ Zero-bias junction capacitance $$\=284.2~\text{fF}\$$.
• $$\\phi=\$$ PB $$\=\$$ VJ $$\=\$$ PN junction potential $$\=0.224~\text{V}\$$.
• $$\m=\$$ MJ $$\=\$$ M $$\=\$$ PN grading coefficient $$\=0.17\$$.
• $$\\tau=\$$ TT $$\=\$$ Transit time $$\=55~\text{ps}\$$.
• $$\V_{br}=\$$ BV $$\=\$$ Reverse breakdown voltage $$\=42~\text{V}\$$.
• $$\E_g=\$$ EG $$\=\$$ Band-gap voltage $$\=0.53~\text{eV}\$$.
• $$\p=\$$ XTI $$\=\$$ IS temperature exponent $$\=1.5\$$.

I found this in c11 of MOSFET Models for VLSI Circuit Simulation:

$$C_j=\frac{C_{j0}}{\left(1-\frac{V_d}{\phi}\right)^m}$$

But he says that this only holds for $$\V_d\leq\frac{\phi}{2}\$$... Another source gives the following:

$$C_j = \cases{\frac{C_{j0}}{\left(1-\frac{V_d}{\phi}\right)^m}, & V_d\leq \text{FC}\cdot\phi\\ \frac{C_{j0}\left(1-\text{FC}(m+1)+\frac{mV_d}{\phi}\right)}{\left(1-\text{FC}\right)^{(m+1)}}, & V_d>\text{FC}\cdot\phi}$$

Is this correct & am I wrong to assume that $$\\text{FC}\$$ (the Forward-bias depletion capacitance coefficient) is always 0.5?